Question

A solenoid consists of 50 turns of wire and has a length of 10 cm. The magnetic field inside the solenoid when it carries a current of 0.5A will be:

Answer 1

3.14 × 10^{-4} T

Answer 2

The magnetic field inside a long solenoid is given by the formula:

$B = \mu_0 n I$

where:

$B$ is the magnetic field strength $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$) $n$ is the number of turns per unit length ($n = N/L$) $I$ is the current flowing through the solenoid

Given values:

Number of turns, $N = 50$ Length of the solenoid, $L = 10 \text{ cm} = 0.1 \text{ m}$ Current, $I = 0.5 \text{ A}$

First, calculate the number of turns per unit length, $n$:

$n = \frac{N}{L} = \frac{50 \text{ turns}}{0.1 \text{ m}} = 500 \text{ turns/m}$

Now, substitute the values into the formula for $B$:

$B = (4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \times (500 \text{ turns/m}) \times (0.5 \text{ A})$ $B = 4\pi \times 10^{-7} \times 250$ $B = 1000\pi \times 10^{-7}$ $B = \pi \times 10^{-4} \text{ T}$

Using the approximate value of $\pi \approx 3.14$:

$B \approx 3.14 \times 10^{-4} \text{ T}$